
For 1-D MCP (\(n=1\) ), \(\mathcal {A}(r,r_\mathrm {d},x)=\min \lbrace r+r_\mathrm {d}-x,2\beta (r)\rbrace \) . Hence, CD distribution is given as
\({1}{-1}&F_{R_\mathrm {C}}(r)=1-\exp \left(-2{\lambda _\mathrm {p}}\left((r+r_\mathrm {d})\vphantom{e^{-jjjj_k}}-|r-r_\mathrm {d}|e^{-\lambda _\mathrm {d}2\beta (r)}+(e^{-\lambda _\mathrm {d}2\beta (r)}-1)/{\lambda _\mathrm {d}}\right)\right).\) 

For the 2-\(n=2\) ) case, \(\mathcal {A}(r,r_\mathrm {d},x)=\)
\({1}{-1} \hspace{-21.68121pt} {\left\lbrace \begin{array}{ll}r_{\mathrm {d}}^2\cos ^{-1}\left(\frac{x^2+r_\mathrm {d}^2- r^2}{2xr_{\mathrm {d}}}\right)+r^2\cos ^{-1}\left(\frac{x^2+ r^2-r_\mathrm {d}^2}{2xr}\right)& {SC}{}{\text{If\,} r+r_\mathrm {d}\ge }\\-\frac{1}{2}\sqrt{((r_\mathrm {d}+r)^2-x^2)(x^2-(r_\mathrm {d}-r)^2)},&\hspace{-7.22743pt}{SC}{\text{if } r+r_\mathrm {d}\ge }{ }\, x \ge |r-r_\mathrm {d}|\\\pi \min (r,r_\mathrm {d})^2=\pi \beta ^2(r)&\text{\,\,\,\,\,\,otherwise}\end{array}\right.}\nonumber \) 

The closed form expression may not be possible for this case. However, the derived CDF for CD (REF ) is significantly simpler than the one in [1]}.

